Lesson 7 · 35 min
Conservation of Energy
Move the work of the weights and springs to the other side of the equation, as potential energy, and the principle of work and energy becomes a statement of conservation: the kinetic energy plus the potential energy stays the same. Friction and other nonconservative forces are what make it change, and they enter as a separate term.
Learning objectives
- Derive \(T_1 + V_1 = T_2 + V_2\) from the principle of work and energy, and say when it holds.
- Use the general form \(T_1 + V_1 + \sum\left(U_{1\to2}\right)_{nc} = T_2 + V_2\) when friction or applied forces do work.
- Count each conservative force once: as work or as potential energy, never both.
- Solve problems with pendulums, launchers, springs whose length changes along the path, and connected bodies.
From work to energy
Split the works in \(T_1 + \sum U_{1\to2} = T_2\) into the conservative ones, weights and springs, and the rest. Lesson 6 showed that each conservative work is a drop in potential energy, \(V_1 - V_2\), with \(V = \colVg{V_g} + \colVe{V_e}\):
\[ T_1 + (V_1 - V_2) + \textstyle\sum\left(U_{1\to2}\right)_{nc} = T_2 \]Move \(V_2\) to the right:
Work and energy with potential energy
\[ \colKE{T_1} + V_1 + \textstyle\sum\left(U_{1\to2}\right)_{nc} = \colKE{T_2} + V_2 \]\(\left(U_{1\to2}\right)_{nc}\) is the work of the nonconservative forces: friction (negative), drag, pushes and pulls, motors. Meriam & Kraige call it \(U'_{1\text{-}2}\).
When only conservative forces do work (forces that do no work, like a smooth surface's push or a pendulum's string, are allowed), the last term vanishes:
Conservation of mechanical energy
\[ \colKE{T_1} + V_1 = \colKE{T_2} + V_2 \]The mechanical energy \(E = T + V\) stays constant: kinetic energy and potential energy trade back and forth, and nothing is lost.
A pendulum
Example 7.1 — Released from horizontal
The bob of Figure 7.1 is released from rest with the string horizontal. How fast is it moving at the bottom of its swing?
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Only the weight does work, so energy is conserved. Put the datum at the bottom, \(L = 1.2\ \text{m}\) below the release point:
\[ 0 + \colVg{mgL} = \tfrac12 mv_2^2 + 0 \quad\Rightarrow\quad v_2 = \sqrt{2gL} = \sqrt{2(9.81)(1.2)} = 4.85\ \text{m/s} \]The mass cancels, and so does the shape of the path, as on the slides of Lesson 1. To find the string's tension at the bottom, you need the normal equation \(\sum F_n = mv^2/\rho\) as well: that is Lesson 8.
Springs and changing geometry
Example 7.2 — A spring launcher
A launcher fires a \(0.2\ \text{kg}\) ball straight up. Its spring, with \(k = 800\ \text{N/m}\), is compressed \(0.1\ \text{m}\) before release. Ignoring air resistance and the spring's mass, how high above its release point does the ball rise?
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Position 1: released, at rest, spring compressed. Position 2: the top, at rest, spring unstretched. Datum at the release point:
\[ 0 + \colVe{\tfrac12(800)(0.1)^2} = 0 + \colVg{(0.2)(9.81)\,h} \quad\Rightarrow\quad h = \frac{4}{1.962} = 2.04\ \text{m} \]The spring's \(4\ \text{J}\) ends up entirely as gravitational potential energy. Over the first \(0.1\ \text{m}\) the spring is still pushing, which is why \(h\) is measured from the compressed position.
Example 7.3 — A collar on a rod, held by a spring
A \(2\ \text{kg}\) collar slides on a smooth vertical rod. A spring with \(k = 200\ \text{N/m}\) and unstretched length \(0.3\ \text{m}\) joins the collar to a fixed point \(C\), \(0.4\ \text{m}\) to the side of the rod. The collar is released from rest at \(A\), level with \(C\). How fast is it moving at \(B\), \(0.3\ \text{m}\) below \(A\)?
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The spring is not along the motion, and its force changes direction as the collar slides: computing its work directly would be hard. Its potential energy only needs its length at each end.
At \(A\): length \(0.4\ \text{m}\), stretch \(s_A = 0.1\ \text{m}\). At \(B\): length \(\sqrt{0.4^2 + 0.3^2} = 0.5\ \text{m}\), stretch \(s_B = 0.2\ \text{m}\). Datum at \(B\). The rod is smooth, so its push does no work:
\[ 0 + \colVe{\tfrac12(200)(0.1)^2} + \colVg{(2)(9.81)(0.3)} = \tfrac12(2)\,v_B^2 + \colVe{\tfrac12(200)(0.2)^2} \] \[ 1 + 5.886 = v_B^2 + 4 \quad\Rightarrow\quad v_B = 1.70\ \text{m/s} \]When friction acts
Example 7.4 — How much did friction take?
A \(60\ \text{kg}\) skier starts from rest and descends \(20\ \text{m}\) vertically, reaching \(15\ \text{m/s}\) at the bottom of the run. How much work did friction and air drag do on the skier?
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Datum at the bottom. The snow's normal force does no work; friction and drag are the nonconservative forces:
\[ 0 + \colVg{(60)(9.81)(20)} + U_{nc} = \tfrac12(60)(15)^2 + 0 \quad\Rightarrow\quad U_{nc} = 6750 - 11\,772 = -5022\ \text{J} \]Friction and drag took \(5.02\ \text{kJ}\), about \(43\%\) of the potential energy the skier started with.
Example 7.5 — Lesson 4's crate, in energy form
Redo part (b) of Example 4.1 (a \(20\ \text{kg}\) crate sliding \(3\ \text{m}\) down a \(30^\circ\) incline with \(\mu_k = 0.25\) into a spring with \(k = 2\ \text{kN/m}\)) with potential energies.
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Datum at the crate's lowest position, at maximum compression \(s\). At the start: \(T = 0\), \(V_g = (98.1)(3 + s)\), \(V_e = 0\). At maximum compression: \(T = 0\), \(V_g = 0\), \(V_e = \tfrac12(2000)s^2\). Friction is the only nonconservative force:
\[ 0 + \colVg{98.1(3 + s)} \colF{{} - 42.48(3 + s)} = 0 + \colVe{\tfrac12(2000)s^2} \]The same quadratic as before, so \(s = 0.437\ \text{m}\). The bookkeeping changed; the physics did not.
Systems of particles
For several particles, add up their kinetic and potential energies: \(\sum T_1 + \sum V_1 + \sum\left(U_{1\to2}\right)_{nc} = \sum T_2 + \sum V_2\). Inextensible cords that stay taut do no net work, as in Lesson 4.
Example 7.6 — A block pulled off a table
Block \(A\) (\(3\ \text{kg}\)) rests on a smooth table and is tied by a cord over a smooth pulley at the edge to block \(B\) (\(2\ \text{kg}\)), which hangs. The system is released from rest. How fast are the blocks moving after \(B\) has fallen \(0.8\ \text{m}\)?
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Only gravity does work on the system (the table is smooth and the cord's tension cancels). \(A\)'s height does not change, so only \(B\)'s potential energy changes. Datum at \(B\)'s final position:
\[ 0 + \colVg{(2)(9.81)(0.8)} = \tfrac12(3 + 2)\,v^2 + 0 \quad\Rightarrow\quad v = \sqrt{\frac{2(15.70)}{5}} = 2.51\ \text{m/s} \]Check your understanding
Key takeaways
- \(T_1 + V_1 + \sum\left(U_{1\to2}\right)_{nc} = T_2 + V_2\), with \(V = V_g + V_e\). With no nonconservative work, \(T_1 + V_1 = T_2 + V_2\).
- Count each weight and spring once: as work or as potential energy.
- Potential energy handles springs whose direction changes along the path: only their length at the two ends matters.
- Friction's work is the drop in mechanical energy; it ends up as heat.
- Next, Lesson 8 uses the speed from energy to find forces on curved paths.